What is the derivative of e^tanx by first principle? Share with your friends Share 9 Varun.Rawat answered this Let y = etan x⇒y + ∆y = etanx+∆x⇒∆y = etanx+∆x - etan x⇒∆y∆x = etanx+∆x - etan x∆x⇒∆y∆x = etan x etanx+∆x - tan x - 1∆x⇒lim∆x→0∆y∆x =etan xlim∆x→0 etanx+∆x - tan x - 1tanx+∆x - tan x × tanx+∆x - tan x∆x⇒dydx = etan x × 1 × lim∆x→0sinx+∆xcosx+∆x - sin xcos x∆x⇒dydx = etan x ×lim∆x→0sinx+∆x . cos x - cosx+∆x . sin x∆x × 1cos x . cosx+∆x⇒dydx =etan x× lim∆x→0sin ∆x∆x × lim∆x→0 1cos x . cosx+∆x⇒dydx = etan x ×1×1cos2x = etan x . sec2x 3 View Full Answer Thundersam answered this . 16